Question

A stone thrown vertically upwards with
initial velocity u reaches a height 'h'
before coming down. Show that the
time taken to go up is same as the time
taken to come down.​

Answer 1

Step-by-step explanation:

When stone throw upward with initial velocity, u

Final velocity at maximum height, v=0

Apply kinematic equation of motion.

v=u+at

t=v-u/a=0-u/-g

Apply kinematic equation of motion.

s=ut+1/2 at²

h=u(u/y)-1/2g(u/g)²

h=u( gu )− 21 g( gu ) ²

u= 2gh

time,

t=u/g=√2gh/g

t=√2h/g......(1)

(ii) When stone drop from height h

Apply kinematic equation of motion.

v²-u²=2gh

v=√2gh

Apply kinematic equation of motion.

v=u+at

t=v-u/a = √2gh-0/g=√2h/g......(2)

From (1 ) and (2) it is proved that the time taken to go up is same as the time taken to come down.

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